10
PASCALE CHARPIN
THEOREM
2. Let x
E
pr and y
E
pm(p- 1)-r . Then the product xy is zero
if and only if Pn(xy)
=
0. Moreover if xy
=
0 then x(y+ pm(p-
1)-r+1)
=
{0}.
Define the polynomial of G[Z]:
R(Z)
=
L
tPi(x)Pn-i(y)zn-i .
wp(i)=r
Then we have for all
j
E
[0,
n-
1] :
(II) Pn(Xaj(y))
=
(-1tR(aJ).
The number of shifts of y (i.e. the aj(y)) which are orthogonal to x equals the
number
of zeros of the polynomial R(Z).
Proof: The product xy is in
prpm(p-1)-r
=
pm(p-1).
A codeword z of
pm(p-1)
is defined by Ps(z)
=
0 unless s
=
n.
If
Pn(z)
=
0, z is the null vector; otherwise
z
=
a1, a E K and 1 is the all-one vector (see Section 2.1).
If
xy
=
0 then
xb
=
0 for every codeword b of the coset y
+
pm(p-1)-r+1,
since xPm(p-
1)-r+1
is in pm(p-
1)+1
which is the null space.
From the definition of the functions Ps and of a1, we have:
Ps(aj(y))
=
,L:
y9(a1g)
8 =
a18
/J8
(y).
gEG
Thus (noting that every i in S satisfies i
-
n):
Pn(Xaj(y))
=
t, (
7 )
tPi(x)Pn-i(aj(y))
=
t, (
7 )
tPi(x)Pn-i(y)a-ji
Since x E pr , Pi(x)
=
0 for all i such that wp(i) r (see Definition 3).
When wp(i) r, wp(n- i) m(p-
1)-
r; in this case Pn-i(Y)
=
0, since
y E pm(p-
1)-r.
Then the sum above is a sum on the i such that wp(i)
=
r.
Now we consider such ani and its p-ary expansion (io, ... , im-1) and we
want to compute the coefficient (
7 )
modulo p:
since ( p
z
1
)
=
(p- 1) ... (p- it)/(it!)
= (
-1)i' (modulo p). Therefore
Pn(xaJ(y))
=
2.:
(-It Pi(x)Pn-i(y)a-Ji
= (
-1r R(aJ) .
wp(i)=r
0
Remarks: 1) The set of the i whose p-weight equals
r
is invariant under the
multiplication by p (modulo n); moreover Psp(z)
=
LgEG
z9 gPs
=
(Ps(z))P, for
any z. The polynomial R(Z) is in fact a trace-function from G to K.
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